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Linear Algebra - Basis

by rad0786
Tags: algebra, basis, linear
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rad0786
#1
Oct4-05, 05:41 PM
P: 188
Hello... im doing this problem with basis. Infact, im having a lot of problems understanding basis, i did every question in the text book and I still get seem to understand the idea of it.

So i was hoping somebody can help me with the whole idea about it.

Say, for example, how would i go abouts a question like this:

--Let F be a field and let V = F^3. Let
W = {(a1 a2 a3) E F^3 / 2a1 - a2 - a3 = 0 }

Find a basis for W --

If a question like that came on a test, id fail it - sad to say. It would also be good if somebody knows a good website or has sameple tests that covers this mataril so that i may get used to it.

Thanks
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Brad Barker
#2
Oct4-05, 05:56 PM
P: 429
so any element in W can be represented like so:

w = (a1, a2, a3), where a1, a2, and a3 are arbitrary.

but W has the additional restriction that a1 = 1/2 (a2 + a3).

so

w= ( 1/2 (a2+a3), a2, a3).

w = a2 ( 1/2, 1, 0) + a3 (1/2, 0, 1). (it's easy to see that this is the same as above.)

so

w = span{(1/2, 1, 0), (1/2, 0, 1)}.


and that set {(1/2, 1, 0), (1/2, 0, 1)} is our basis.


ah, i miss these problems!
Gokul43201
#3
Oct4-05, 06:04 PM
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Keep in mind that the above is not the only basis.

Also, notice that the given vector space is nothing but a (generalization of a) plane through the origin in [itex]\mathbb{R}^3[/itex]. Any pair of vectors in the plane will serve as a basis.

rad0786
#4
Oct4-05, 09:06 PM
P: 188
Linear Algebra - Basis

Hey.. i was just wondering about Brad Barkers post above...

he said that a1 = 1/2 (a2 + a3).

well.. shouldn't it be a1 = 1/2 (a2 - a3) ?

Does it make a difference?
Gokul43201
#5
Oct5-05, 01:06 AM
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No, you said "2a1 - a2 -a3 = 0"

That gives 2a1 = a2 + a3
rad0786
#6
Oct5-05, 04:49 AM
P: 188
oh that was my silly mistake.. but either way.. i still learned something :)
rad0786
#7
Oct5-05, 11:16 AM
P: 188
Hey.... what about the subspace...

U = (a+b+c=0/ a b c is in the Real Numbers)

How would you show the span of that.

Also, the comment Gokul43201 made, about the basis being the plane through the origin, how did he know that? I mean its a plane because of the two vectors, but how did he know that its through the origin?
iNCREDiBLE
#8
Oct5-05, 02:54 PM
P: 128
Quote Quote by rad0786
Also, the comment Gokul43201 made, about the basis being the plane through the origin, how did he know that? I mean its a plane because of the two vectors, but how did he know that its through the origin?
The general equation of a plane is [itex]Ax+By+Cz = D[/itex].
[itex]D = 0 \Longleftrightarrow[/itex] the plane goes through the origin.
rad0786
#9
Oct5-05, 05:36 PM
P: 188
"and that set {(1/2, 1, 0), (1/2, 0, 1)} is our basis." Can that be a Basis for F^3? what is F (Feild)? isnt that the same as R, like R^3
Gokul43201
#10
Oct6-05, 01:44 AM
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Quote Quote by iNCREDiBLE
The general equation of a plane is [itex]Ax+By+Cz = D[/itex].
[itex]D = 0 \Longleftrightarrow[/itex] the plane goes through the origin.
Also, for a plane to constitute a vector space with the usual vector addition, it must pass through the origin, since this point is the additive identity element.


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